**Basic Concepts**

**Basic Concepts**

* Sequence: *Some numbers arranged in a definite order, according to a definite rule, are said to form a sequence.

eg. (i) 1,2,3,………….

(ii) 100, 60, 30, ……………

(iii) -1.0, -1.5, -2.0, -2.5, ……….

**Arithmetic Progressions(AP): **

An arithmetic progression is a list of numbers on which each term is obtained by adding a fixed number to the preceding term except the first term.

→ This fixed number called Common difference of A.P.

→ Common difference can be positive, negative or zero.

** Exercise 5.1**

**Q.1 In which of the following situations, does the list of numbers involved make an arithmetic progression, and why?**

**(i) The taxi fare after each km when the fare is Rs 15 for the first km and Rs 8 for each additional** **km.**

* Solution*: Yes it is in A.P.

Because 15, 13, 31, ………….. each succeeding term is obtained by adding 8 in its preceding term.

**(ii)The Amount of air present in a cylinder when a vaccum pump removes 1/4 of the additional km.**

**Solution:**

No it is not in A.P. Volume v, 3/4 v , (3/4)² v

**(iii) The cost of digging a well after every metre of digging, when it costs Rs 150 for each subsequent metre the first metre and rises by Rs 50**

* Solution: *yes it is in A.P.

Because 150, 200, 250, ….each succeeding term is obtained by adding 50 in its preceding.

(iv) The amount of money in the account every year, when Rs 10000 is deposited at compound interest at 8% per annum.

**Q.2 Write first four terms of the A.P. , When the first term a and the common difference ‘d’ are given as follows:**

**(i) a = 10, d = 10**

**Solution:**

a_{2} = a+d = 10 +10 = 20

a_{3} = a2 + d = 20 +10 = 30

a_{4} = a3 + d = 30 + 10 = 40

a_{5} = a4 + d = 40 + 10 = 50

Hence first four terms are 10,20,30,40

**(ii) a = -2, d = 0**

**Solution:**

a_{2} = a + d = -2 + 0 = -2

a_{3} = a2 + d = -2 + 0 = -2

a_{4} = a3 + d = -2 + 0 = -2

a_{5 }= a4 + d = -2 + 0 = -2

Hence first four terms are -2,-2,-2,-2

**(iii) a = 4, d=-3**

**Solution:**

a_{2} = a + d = 4-3 = 1

a_{3} = a_{2} + d = 1-3=-2

a_{4 }=a_{3}+ d = -2-3 = -5

Hence first four terms are 4,-3,1,-2,-5

**(iv) a = -1, d = 1/2**

**Solution:**

a_{2 }= a + d = -1 + 1/2 = -2+1/2 = -1/2

a_{3 }= a_{2}+d = -1/2 + 1/2 = 0

a_{4 }= a_{3}+d = 0 + 1/2 = 1/2

Hence first four terms are 1, 1/2, -1/2, 0, 1/2

**(v) a = -1.25, d = -0.25**

**Solution:**

a_{2 }= a + d = -1.25 – 0.25 = -1.50

a_{3}=a_{2}+d = -1.50 – 0.25 = -1.75

a_{4 }= a_{2}+ d = -1.75 – 0.25 = -2.0

Hence the four terms are -1.25, -1.50, -1.75, -2.0

**Q.3 For the following APs, write the first term and the common difference:**

**(i) 3,1,-1,-3,…….**

**Solution:**

First term (a) = 3

Common difference (d) = 1-3=-2

**(ii) -5,-1,3,7,…….**

**Solution:**

First term (a) = -5

Common difference (d) = -1-(-5)= -1+5 = 4

**(iii) 1/3,5/3,9/3,13/3,………….**

**Solution:**

First term (a) = 1/3

Common difference (d) = 5/3 -1/3 = 4/3

**(iv) 0.6,1.7,2.8,3.9,…….**

**Solution:**

First term(a) = 0.6

Common difference(d) = 1.7 – 0.6 = 1.1

**Q,4 Which of the following are APs? If they form an AP, find the common difference d and write three more terms.**

**(i) 2,4,8,16,……**

**Solution:**

d_{1}= 4-2 = 2

d_{2} = 8 – 4 = 4

d_{3} = 16 – 8 = 8

here d_{1}≠ d_{2}≠ d_{3}

Hence it is not in AP

**(ii) 2, 5/2, 3, 7/2…….**

**Solution:**

First term a = 2

d_{1 }= 5/2 – 2 = (5-4)/2 = 1/2

d_{2} = 3 – 5/2 = (6-5)/2 = 1/2

d_{3 }= 7/2 – 3 = (7-6)/2 = 1/2

d_{1}= d_{2}= d_{3}= 1/2

Hence it is in AP

Common difference (d) = 1/2

Next three terms:

d_{5 }= d_{4}+d = 7/2 + 1/2 = 8/2 = 4

d6 = d_{5}+d = 7+1/2 = 15/2

d_{7 }= d_{6 }+d = 15/2 + 1/2 = 16/2 = 8

**(iii) -1.2, -3.2, -5.2, -7.2, ….**

**Solution:**

First term (a) = -1.2

d_{1 }= -3.2 -(-1.2) = -3.2 + 1.2 = -2.0

d_{2 }= -5.2 – (-3.2) = -5.2 + 3.2 = -2.0

d_{3}= -7.2 – (-5.2) = -7.2 + 5.2 = -2.0

d_{1}= d_{2}= d_{3}= -2.0

Yes it in A.P.

Next three terms:

d_{5 }= d_{4}+d = -7.2 -2.0 = -9.0

d6 = d_{5}+d = -9.0 – 2.0 = -11.0

d_{7 }= d_{6 }+d = -11.0 – 2.0 = -13.0

**(iv) -10, -6, -2, 2 ,……**

**Solution:**

First term (a) = -10

d_{1 }= -6 – (-10) = -6 + 10 = 4

d_{2 }= -2 – (-6) = -2 + 6 = 4

d_{3 }= 2 -(-2) = 2+2 = 4

d_{1}= d_{2}= d_{3}= 4

Yes it is in A.P.

Common difference = d = 4

Next three terms:

d_{5 }= d_{4}+d = 2+4 = 6

d6 = d_{5}+d = 6+4 = 10

d_{7 }= d_{6 }+d = 10+4 = 14

**(v) 3, 3+√2, 3+2√2, 3+3√2, …….**

**Solution:**

d_{1 }= 3+√2 – 3 = √2

d_{2 }= 3+2√2 -( 3+√2 ) = 3 + 2√2 -3 -√2 =√2

d_{3 }= 3+3√2 – (3+2√2) = 3+3√2 – 3 – 2√2 = √2

d_{1}= d_{2}= d_{3}= √2

Yes it is in A.P.

Common difference (d) = √2

Next three terms:

d_{5 }= d_{4}+d = 3+3√2 + √2 = 3 + 4√2

d6 = d_{5}+d = 3 + 4√2 + √2 = 3 + 5√2

d_{7 }= d_{6 }+d = 3 + 5√2 + √2 = 3 + 6√2

**(vi) 0.2, 0.22, 0.222, 0.2222, ……….**

**Solution:**

First term a = 0.2

d_{1 }=0.22 – 0.2 = 0.02

d_{2 }= 0.222 – 0.22 = 0.002

d_{1}≠d_{2}

No it is not in A.P.

**(vii) 0, -4, -8, -12,…….**

**Solution:**

First term = 0

d_{1}= -4-0 = -4

d_{2 }= -8 -(-4)= -8 + 4 =-4

d_{3 }= -12-(-8) = -12+8 =-4

d_{1}=d_{2}=d_{3}= -4

Common difference(d) = -4

Next three term:

d_{5 }= d_{4}+d = -12-4=-12-4=-16

d6 = d_{5}+d = -16 -4 = -20

d_{7 }= d_{6 }+d = -20 – 4 = -24

**(viii) -1/2, -1/2, -1/2, -1/2, ……….**

**Solution:**

First term = a = -1/2

d_{1}= -1/2 -(-1/2) = -1/2 + 1/2 = 0

d_{2 }= -1/2 -(-1/2) = -1/2+1/2 = 0

d_{3 }= -1/2 -(-1/2) = -1/2 +1/2 = 0

d_{1 }= d_{2 }= d_{3 }= 0

Yes it is in A.P.

Next three terms:

-1/2, -1/2, -1/2

**(ix) 1, 3, 9, 27, ……**

**Solution:**

First term (a) = 1

d_{1}= 3-1 = 2

d_{2 }= 9-3 = 6

d_{1}≠d_{2}

No it is not in A.P.

**(x) a, 2a, 3a, 4a, ……**

**Solution:**

First term = a

d_{1}= 2a – a = a

d_{2 }= 3a – 2a = a

d_{3 }= 4a – 3a = a

d_{1}=d_{2}=d_{3}= a

Yes it is in A.P.

Next three terms: 5a, 6a, 7a

**(xi) a, a², a³, a ^{4 }, ……**

**Solution:**

First term =a

d_{1}= a² – a = a(a-1)

d_{2 }= a³ – a² = a²(a-1)

d_{1}≠d_{2}

No it is not in A.P.

**(xii) √2, √8, √18, √32, ……..**

**Solution:**

Firs term = a = √2

d_{1}= √8 -√2 = 2√2 – √2 = √2

d_{2 }= √18 – √8 = 3√2 – 2√2 = √2

d_{3 }= √32 – √18 = 4√2 – 3√2 = √2

d_{1}=d_{2}=d_{3}= √2

Yes it is in A.P.

Common difference = √2

Next three terms:

d_{5 }= d_{4}+d = √32 + √2 = √64 = 8

d6 = d_{5}+d = √64 + √2 = √128

d_{7 }= d_{6 }+d = √128 + √2 = √256 = 16

**(xiii) √3,√6,√9,√12,…….**

**Solution:**

First term (a) = √3

d_{1}= √6 -√3

d_{2}= √9 -√6

d_{1}≠d_{2}

No it is not in A.P.

**(xiv) 1²,2²,3²,7²,….**

**Solution:**

First term (a) = 1²

d_{1 }= 2² – 1² = 4 – 1 = 3

d_{2 }= 3² – 2² = 9-4 = 5

d_{1}≠d_{2}

No it is not in A.P.

**(xv) 1²,5²,7²,73,…..**

**Solution:**

First term = 1²

d_{1 }= 5² – 1² = 25-1 = 24

d_{2 }= 7² – 5² = 49 – 25 = 24

d3_{ }= 73 – 7² = 73 – 49 = 24

d_{1}=d_{2}=d_{3}= 24

Yes it is in A.P.

Next three terms

d_{5 }= d_{4}+d = 73+24 = 97

d6 = d_{5}+d = 97+24 = 121 = 11²

d_{7 }= d_{6 }+d = 121 + 24 = 145